Piece of trivia for you: The coefficient of the t^3 term is constant. This means that the above equation for displacement is written in terms of a constant "jerk." (No kidding! The jerk, j, is defined as the first time derivative of the acceleration.)

We are told that the stone hits the ground at -165 ft/sec.When does this happen?. . It happens when $\displaystyle y(t) = 0$
So we have: .$\displaystyle h_o - 16t^2\:=\:0\quad\Rightarrow\quad t = \frac{\sqrt{h_o}}{4}$ seconds.

Velocity is the derivative of position: .$\displaystyle v(t)\:=\:-32t$

At that time, we have: .$\displaystyle -32t \:=\:-165\quad\Rightarrow\quad-32\left(\frac{\sqrt{h_o}}{4}\right)\:=\:-165$